Table of Contents
Defining capacitance
Consider two parallel conducting plates of area $A$ separated by a distance $d$. Such a device is called a capacitor. If we connect each plate to a voltage source like a battery or power supply, a charge will appear on each plate, as shown in Fig. 3.
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| Figure 3: Parallel plate capacitor connected to a voltage source |
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The amount of charge $q$ held in a capacitor depends on the capacitor’s capacitance $C$ and on the voltage $V$ applied as
| $ q = CV$ | $(1)$ |
It can be shown that the capacitance is given by
| $C = \dfrac{\epsilon_0 A}{d}$, | $(2)$ |
where $\epsilon_0 = 8.85418782\times 10^{-12}\,\,\dfrac{\mathrm{s}^4\mathrm{A}^2}{\mathrm{m}^3 \,\mathrm{kg}}$ is the permittivity of free space.
The SI unit of capacitance is the farad. A capacitor is said to have a capacitance of one farad when one coulomb of charge produces a potential difference of one volt. The farad is an enormous unit, however. It is more typical to find capacitors on the order of pico-, nano-, or microfarads.
Discharging a capacitor
If a resistor is connected across a charged capacitor by closing a switch as shown in Fig. 4, the excess electrons on the negative plate will move through the resistor to the positively charged (electron-deficient) plate. In other words, the capacitor will discharge.
Figure 4: Capacitor discharging circuit
Using Kirchhoff's rules, the voltage across the capacitor – which by $(1)$ is $V = q/C$ – must be equal to the voltage across the resistor – which is $V=IR$. Therefore,
| $\dfrac{q}{C} = IR$. | $(3)$ |
Since $q$ and $I$ are related by the equation
| $I = -\dfrac{dq}{dt}$, | $(4)$ |
we can substitute $(4)$ into $(3)$ and rearrange to find
| $\dfrac{1}{q}dq = -\dfrac{1}{RC}dt$. | $(5)$ |
Integrating this and putting in the initial condition $q(0) = q_0$, we find
| $\displaystyle q(t) = q_0 e^{-t/RC}$, | $(6)$ |
or, since by $(1)$, $q_0 = C V_0$ and $q=CV$,
| $\displaystyle V_C(t) = V_0 e^{-t/RC}$. | $(7)$ |
Thus, the voltage across the capacitor (or resistor) must decay exponentially. The time for the voltage to drop to $1/e \approx 37\%$ of its initial value will be when $t/RC = 1$, i.e., when $t=RC$. We define this time to be the time constant of a resistance-capacitance or “$RC$” circuit: $\tau = RC$
A graph showing the prediction of $(7)$ for a discharging capacitor is shown in Fig. 5.
Figure 5: Discharge of a capacitor
Charging a capacitor
Suppose instead of starting with the capacitor charged, we start with it discharged and proceed to charge the capacitor through a resistor using a circuit such as that shown in Fig. 6.
Figure 6: Capacitor charging circuit
Using Kirchhoff's rules, $(1)$, and Ohm's law, the equation of the circuit is
| $V_0 = IR + \dfrac{q}{C}$. | $(8)$ |
Substituting $(4)$ for $I$, integrating, and putting in the boundary conditions, we find
| $q(t) = CV_0 \left(1-e^{-t/RC}\right)$, | $(9)$ |
where $V_0$ is the voltage of the battery. In terms of the voltage across the capacitor,
| $V_C(t) = V_0\left(1-e^{-t/RC}\right)$. | $(10)$ |
A graph showing the charging of a capacitor according to the predictions of $(10)$ is shown in Fig. 7. Again, the time constant $\tau= RC$ has meaning. This time, it is the time to reach $1-e^{-1} \approx 63\%$ of its final charge.
Figure 7: Voltage vs. time for a charging capacitor